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algorithm & bitwise operation & the best leetcode solutions

leetcode 136 single-number

the better solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var singleNumber = function(nums) {
  return nums.reduce((sum, i) => sum ^ i, 0);
};
// Time complexity : O(n)
// Space complexity : O(1)

my solution

solution 1

"use strict";

/**
 *
 * @author xgqfrms
 * @license MIT
 * @copyright xgqfrms
 * @created 2020-08-015
 * @modified
 *
 * @description 136 single-number
 * @difficulty Easy
 * @complexity O(n)
 * @augments
 * @example
 * @link
 * @solutions
 *
 */

const log = console.log;

var singleNumber = function(nums) {
  let len = nums.length;
  // obj unique key
  const obj = {};
  while(len) {
    const value = nums[len - 1];
    if(obj[value] === undefined) {
      obj[value] = 1;
    } else {
      obj[value] += 1;
    }
    len--;
  }
  // log(`
keys`, Object.keys(obj))
  // log(`values`, Object.values(obj))
  // log(`entries`, Object.entries(obj))
  for (const key in obj) {
    if (obj.hasOwnProperty(key)) {
      const item = obj[`${key}`];
      if(item === 1) {
        return key;
      }
    }
  }
};
// Time complexity : O(n)
// Space complexity : O(n)

/*

输入: [2,2,1]
输出: 1

输入: [4,1,2,1,2]
输出: 4

*/

const test = [2,2,1];
const result = singleNumber(test);
log(`result =`, result)
// 1

const test2 = [4,1,2,1,2];
const result2 = singleNumber(test2);
log(`result2 =`, result2)
// 4

如何使用 js 计算两个数组的交集、差集、并集、补集

https://www.hangge.com/blog/cache/detail_1862.html

refs

https://leetcode-cn.com/problems/single-number/

https://leetcode.com/problems/single-number/

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